Given the following information at 25C, calculate �G at 25 degrees Celsius
2A(g) � B(g) ±£ 3C(g)
Substance �Hf � (kJ/mol) S�(J/molK)
A(g) 191 244
B(g) 70.8 300
C(g) �-197 164
a. -�956 kJ b. 956 kJ c. �346 kJ
d. 346 kJ e. -�1.03 �* 10^3 kJ
Using dH and dS calculate dGf for A, B, C, then
dGrxn = (n*dGf products) - (n(dGf reactants)
thank you
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Using dH and dS calculate dGf for A, B, C, then
dGrxn = (n*dGf products) - (n*dGf reactants)
To calculate the Gibbs free energy change (ΔG) at 25 degrees Celsius (298 K), we can use the equation:
ΔG = ΔH - TΔS
Where:
ΔH is the change in enthalpy
ΔS is the change in entropy
T is the temperature in Kelvin
Now, let's calculate the values for each component:
For the reaction: 2A(g) → B(g) + 3C(g)
ΔH = [Σ(Enthalpy of Products)] - [Σ(Enthalpy of Reactants)]
ΔH = [1 * ΔHf(B)] + [3 * ΔHf(C)] - [2 * ΔHf(A)]
Substituting the given values:
ΔH = (1 * 70.8 kJ/mol) + (3 * -197 kJ/mol) - (2 * 191 kJ/mol)
ΔH = 70.8 kJ/mol - 591 kJ/mol - 382 kJ/mol
ΔH = -902.2 kJ/mol
ΔS = [Σ(Entropy of Products)] - [Σ(Entropy of Reactants)]
ΔS = [1 * S(B)] + [3 * S(C)] - [2 * S(A)]
Substituting the given values:
ΔS = (1 * 300 J/molK) + (3 * 164 J/molK) - (2 * 244 J/molK)
ΔS = 300 J/molK + 492 J/molK - 488 J/molK
ΔS = 304 J/molK
Now we have both ΔH and ΔS, and the temperature T is given as 25 degrees Celsius, which is 298 K.
Let's substitute these values into the equation:
ΔG = ΔH - TΔS
ΔG = -902.2 kJ/mol - (298 K * 0.304 kJ/molK)
Calculating the multiplication: 298 K * 0.304 kJ/molK = 90.592 kJ/mol
Now subtracting: ΔG = -902.2 kJ/mol - 90.592 kJ/mol
ΔG = -992.792 kJ/mol
So the correct answer is:
a. -956 kJ